Calculating Average Using Bitwise Operators

To be direct the simple average in math is $(x+y)/2$ but we can also do it using

$(x\&y)+\frac{x\text{textasciicircum}y}{2}$

now lets try to prove it

$x+y=2(x\&y)+x\text{textasciicircum}y$ If you forgot bitwise operators

1. x & y = 1s in both x and y
2. x^y = 1s exclusively in x and y ( contains both 1 and 0 )

Lets find out x which is 1s in both x and y + 1s in x